Question

The relation $f$ is defined by $f\left(x\right)=\{\begin{array}{cc}{x}^2,0\le x\le 3& \\ 3x,3\le x\le 10& \end{array}$
The relation $g$ is defined by $g\left(x\right)=\{\begin{array}{cc}{x}^2,0\le x\le 2& \\ 3x,2\le x\le 10& \end{array}$

Show that $f$ is a function and $g$ is not a function.

Answer 1

The relation $f$ is defined as $f\left(x\right)=\{\begin{array}{cc}{x}^2,0\le x\le 3& \\ 3x,3\le x\le 10& \end{array}$

It is observed that for $0\le x\le 3,$ we have $f\left(x\right)=x^2$ and for $3\le x\le 10,$ we have $f\left(x\right)=3x$

Also at $x=3$, $f\left(x\right)=3^2=9$ or $f\left(x\right)=3\times 3=9$

i.e., at $x=3,f\left(x\right)=9$

Therefore for every $x,$ $0\le x\le 10$, we have unique image under $f$

Thus, the relation $f$ is a function.
Also, the relation $g$ is defined as $g\left(x\right)=\{\begin{array}{cc}{x}^2,0\le x\le 2& \\ 3x,2\le x\le 10& \end{array}$

It can be observed that for $x=2,$ we have $g\left(x\right)=2^2=4$ and $g\left(x\right)=3\times 2=6$

Thus, the element $2$ of the domain of the relation $g$ has two different images i.e., $4$ and $6$

Hence, this relation is but not a function.